Problem: Find $\dfrac{d}{dx}\left[3^{(2x-x^3)}\right]$. Choose 1 answer: Choose 1 answer: (Choice A) A $\ln(3)\cdot 3^{(2x-x^3)}$ (Choice B) B $\ln(2-3x^2)\cdot 3^{(2x-x^3)}$ (Choice C) C $\ln(3)\cdot 3^{(2-3x^2)}$ (Choice D) D $\ln(3)\cdot 3^{(2x-x^3)}\cdot (2-3x^2)$
Explanation: $3^{(2x-x^3)}$ is an exponential function, but its argument isn't simply $x$. Therefore, it's a composite function. In other words, suppose $u(x)=2x-x^3$, then $3^{(2x-x^3)}=3^{u(x)}$. $\dfrac{d}{dx}3^{(2x-x^3)}$ can be found using the following identity: $\dfrac{d}{dx}\left[3^{u(x)}\right]=\ln(3)\cdot 3^{u(x)}\cdot u'(x)$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}3^{(2x-x^3)} \\\\ &=\dfrac{d}{dx}3^{u(x)}&&\gray{\text{Let }u(x)=2x-x^3} \\\\ &=\ln(3)\cdot 3^{u(x)}\cdot u'(x) \\\\ &=\ln(3)\cdot 3^{(2x-x^3)}\cdot (2-3x^2)&&\gray{\text{Substitute }u(x)\text{ back}} \end{aligned}$ In conclusion, $\dfrac{d}{dx}3^{(2x-x^3)}=\ln(3)\cdot 3^{(2x-x^3)}\cdot (2-3x^2)$